Mathematics - 2014-08-18 (page 1 of 2)

900 sit-ups a day

00:46

:( Was it something I said? math.stackexchange.com/users/61527/t-bongers

Kaj Hansen

@TedShifrin, ready for probability?

:P

Mike Miller

@900sit-upsaday Any number of lost users is better than MSE continuing its descent into a repository for infinitely-repeated homework questions.

3

r9m

01:02

@Chris'ssis Not yet :(((

blue

01:18

man, pparently only one other soul appreciated the arxiv joke I posted :(

Mike Miller

I snickered

silently

skullpatrol

02:08

Like you said, it's a sophisticated joke

Try the Homotopy chat room.

Mike Miller

03:50

this was 2.5 years ago...

skullpatrol

...nvm

Babak S.

04:53

Hi there! :-)

Martin Sleziak

I wondered whether removing homework will somehow bring new batch of generalist badges. It seems that algebraic-topology and commutative-algebra have similar numbers of questions and probably sometimes one of them is ahead (and in top 40 tags) and sometimes the other one. Removal of homework tag will put both these tags into top 40 tags.

Babak S.

@MartinSleziak: Good time! Is there any reference in which we can have almost all about Trigonometry ? I couldn't find such this on the site. Thanks

Martin Sleziak

What "almost all" means.

List of trigonometric identities at Wikipedia?

Good book recommendations on trigonometry

Babak S.

@MartinSleziak: I mean such a book , a famous one.

Martin Sleziak

I don't know what you are looking for. (But I am not particularly knowledgeable of trigonometry.)

Babak S.

05:02

Thanks. I got the second link. Thanks.

Martin Sleziak

Hans-Jochen Bartsch?

Babak S.

@MartinSleziak: May I ask you: Do you know Dr. M. Eshagi at Semnan university?

Martin Sleziak

No, I don't.

Maybe you will ask something useful among questions tagged trigonometry+reference-request.

Babak S.

Thanks again.

Sush

06:12

Team 900 will make math.se an empty castle! Or, someday, I will visit this site only for the same reasons as I do visit the EJMR.

r9m

06:24

@Sush what is EJMR ?

The Substitute

I am reading notes on analysis that proves (1) Every bounded sequence in the reals has a subsequence that converges to the supremum of the sequence and (2) The Bolzano-Weierstrass Theorem, which says that any bounded sequence has a convergent subsequence. Is there any practical use in treating (1) and (2) as if they are not equivalent?

Jasper Loy

06:56

@blue On the other hand, nobody appreciated the gif I posted.

eXtremiity

@TheSubstitute.
(1) says that a bounded sequence has a subsequence that converges TO THE supremum of the original sequence. That is, (1) says where the subsequence will converge to.

(2) says that given any bounded sequence, it has a subsequence that will converge. Furthermore, the original sequence can be bounded and DIVERGENT, but will still contain a subsequence that converges. However, the theorem does not indicate where it converges to.

Hold on, I'm having second thoughts. (1) can be a bounded divergent sequence, will still have a supremum, and therefore the subsequence will converge to it. Hmm.

Martin Sleziak

07:20

It's virtually impossible to keep the pace with 900 sit-ups a day in the number of the votes cast. He leads this week, month, quarter and year.

Sush

Please let me know why $\sum_{j=1}^n|x_j|\le\sqrt{n\cdot\sum_{j=1}^nx_j^2}$?

@MartinSleziak

Martin Sleziak

@Sush Isn't that just Cauchy-Schwarz, when one of the vectors is (1,1,...1).

Sush

@MartinSleziak, ok, let me see!

Martin Sleziak

BTW it seems to be basically this question: Why is this true: $\|x\|_1 \le \sqrt n \cdot \|x\|_2$?

I have to leave. See you later!

skullpatrol

Later

Sush

07:28

@MartinSleziak, thanks for making my mind walk little forward.

@skullpatrol, hi!

skullpatrol

Hi pal :)

Sush

@DanielFischer, thank you so much for helping me last night.

Daniel Fischer

You're welcome, @Sush.

Dobry den, @MartinSleziak.

skullpatrol

Has the excitement over winning the World Cup settled down yet @Daniel?

Daniel Fischer

@skullpatrol I never won any world cup, as far as I remember.

The excitement is about to start, because next weekend the season is going underway, @skullpatrol.

skullpatrol

07:39

What season is that?

Daniel Fischer

@skullpatrol The league, football/soccer.

skullpatrol

Well, being world champions makes it that much more special, no?

Sush

@DanielFischer, my book has Cauch-Schwartz as $|x\cdot y|\le \|x\|\|y\|$, where $x$ and $y$ are vectors. So, letting $y=1$, we get $|\sum_{j=1}^nx_j|\le\sqrt{n\cdot\sum_{j=1}^nx_j^2}.$ But triangular inequality gives $|\sum_{j=1}^nx_j|\leq\sum_{j=1}^n|x_j|,$ so how is $\sum_{j=1}^n|x_j|\le\sqrt{n\cdot\sum_{j=1}^nx_j^2}$ guaranteed?

blue

do CS with (|x_1|,...,|x_n|)

Daniel Fischer

@Sush Consider $\tilde{x} = (\lvert x_1\rvert, \lvert x_2\rvert,\dotsc,\lvert x_n\rvert)$. Since $x_k^2 = \lvert x_k\rvert^2$ ...

And, @Sush, Schwarz without 't', it's Hermann Amandus, not Laurent.

Sush

07:45

@DanielFischer, ok! sorry for mistake!

Daniel Fischer

@Sush No problem. A lot of people mix them up.

Sush

@DanielFischer, thank you so much!!! I didn't know that such little things as $x_k^2 = \lvert x_k\rvert^2$ can be so useful!

Daniel Fischer

@Sush The little things are immensely useful. One just doesn't notice it so often because they are so little.

Sush

@DanielFischer, I just saw "A Beautiful Mind". Did Nash really stuck with those people of his past? Did he work for Pentagon as code breaker? I have doubt because some sources say that he did not give a speech while receiving Nobel Prize. So, there might be other things just to make movie beautiful, I think.

Daniel Fischer

@Sush I know pretty little of Nash. His embedding theorem, of course. And that he was schizophrenic (if that is in fact the correct diagnosis, I'm not sure about that). Beyond that, not so much.

Sush

07:56

@DanielFischer, ok, thank you so much!

Alizter

08:14

@Chris'sis Prove that $$\lim_{n\to\infty} \prod^n_{i=0}\Gamma\left(1+\frac in\right)^{\Gamma\left(1+\frac in\right)/n}=e$$

Jasper Loy

@robjohn It's late, you should be sleeping.

@Sush The real life is quite different from the movie. Read the book to find out. But the movie is nonetheless inspiring.

Jasper Loy

08:39

@900sit-upsaday I have deleted my account about ten times, don't worry about it.

Alizter

08:53

@Chris'ssis above sorry

Chris's sis

09:26

Greetings

@Alizter Does it tend to $e$? How is that possible?

@Alizter Question: what is the first thing to note? Don't hurry, think of it.

Alizter

@Chris'ssis I have a beautiful proof for it :)

Chris's sis

@Alizter Teach me.

Alizter

Start with ze log

Chris's sis

OK ... and then ...

Alizter

It turns into a riemann sum

Chris's sis

09:38

@Alizter Yeah, that's clear.

Alizter

So then the intragral is $\int_1^2 \log{\Gamma(X)}^{\Gamma(x)}dx$

rehband

Hey guys, I have a newbie question here: How do I find the partial derivatives of $$h(x_1,..,x_n)=\int_{0}^{||x||} f(t) dt$$ where $||x||$ is the Euclidean norm of $x=(x_1,...,x_n)$ and $f$ is some continuous function?

Chris's sis

@Alizter and then?

Alizter

Hmm I may have done something wrong here

Chris's sis

@Alizter Well, I'm glad you see that. :-)

Alizter

09:42

wait no

Chris's sis

Digamma function

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: It is the first of the polygamma functions. == Relation to harmonic numbers == The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as == Integral representations == If the real part of x is positive then the digamma function has the following integral representation...

Alizter

@Chris'ssis Yes

I have a mistake ooops

Chris's sis

@Alizter This one would have been a nice variant though $$\lim_{n\to\infty} \prod^n_{i=0}\Gamma\left(1+\frac in\right)^{1/n}$$

OR

$$\lim_{n\to\infty} \prod^n_{i=0}\Gamma\left(1+\frac in\right)^{(1+i/n)/n}=e^{\large\frac{3}{4}\log(2\pi)-\log(A)-\frac{5}{4}}$$

Alizter

I don't know what I was thinking :)

Chris's sis

@Alizter try this one - Prove that

$$\sum_{n=1}^{\infty}\left(\psi^{(0)}(n)+\frac{3}{2}\psi^{(1)}(n)+\frac{3}{2}\psi^{(2)}(n)+\frac{1}{2}\psi^{(3)}(n)+\frac{1-n}{n^2}-\log(n)\right)$$
$$=2+\frac{\gamma}{2}-\frac{\log(2\pi)}{2}-2\zeta(2)+3\zeta(3)$$

Actually I'm only interested to tell me the first step to do. Take your time! :-)

This is about the art of solving problems.

I'm sure many would do it with the risk of getting a messy proof. The point is to get a marvellous proof.

Mew

10:04

hi

anyone in?

Jasper Loy

I am here, but I cannot help you with math.

Mew

aw

pretty simple question tho

how to solve tan(x) = sqrt(2) - 1

i know the answer is pi/8, but what is the easiest way to get it

Chris's sis

$$\tan(a/2)=\sin(a)/(1+\cos(a))$$

$$ \sqrt{2} - 1=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}/2}{1+\sqrt{2}/2}=\frac{\sin(\pi/4)}{1+\cos‌(\pi/4)}=\tan(\pi/8)$$

Jasper Loy

@Chris'ssis Wow, you are a genius!

Jasper Loy

10:38

@MikeMiller Although I agree with your comment about spamming others, I don't think people will necessarily look at his question.

Alizter

@Chris'ssis I start with $\sum_{k=n}^\infty\frac1{k^{m+1}}$

Mew

ty Chris's sis

robjohn

11:47

@Chris'ssis I assume you have one...

Jasper Loy

@robjohn You are up early, lol.

Chris's sis

@robjohn Sure. Welcome back! (btw) :-)

robjohn

@Chris'ssis I was worried about the $+\frac32\psi'(n)$ term until I noticed the $-\frac1n$ term later

Jasper Loy

@Chris'ssis What is the distance you run every day?

Chris's sis

@JasperLoy approx. 3-4 km

Jasper Loy

11:51

@Chris'ssis Very impressive.

Chris's sis

@JasperLoy It really makes you feel better, especially mentally. :-)

@robjohn You have a solution, right?

robjohn

@Chris'ssis I just saw the question. I know the asymptotics, so I know whether it converges.

Chris's sis

@robjohn That's a very good starting point.

Jasper Loy

@robjohn Are you familiar with Banach manifolds or integrals of Banach-valued functions?

Chris's sis

@Alizter see what @robjohn said above.

Maths Lover

11:57

Hello

Jasper Loy

I ask because Lang uses Banach manifolds in his "Fundamentals of differential geometry" and integrals of Banach-valued functions in his "Real and functional analysis", but it seems they are not mainstream @robjohn.

Maths Lover

Is there anyone familiar with Munkres topology?

Jasper Loy

@MathsLover What do you want to ask?

Maths Lover

I wonder If metric spaces(two sections of chapter 2) are required to study chapter 3( about connectedness and compactness)

@JasDer

@JasperLoy

Jasper Loy

@MathsLover I can't tell you offhand like that, but metric spaces are very important, so you should study them.

Maths Lover

12:00

I've already studied them but quickly, Not in depth though.

Jasper Loy

@MathsLover OK, then I guess you can move on to the next chapter first, though it's always good to study books from cover to cover.

Maths Lover

Why is that?

@JasperLoy

I mean, why is it better to study books from cover to cover?

Jasper Loy

@MathsLover To get the most information out of it?

Maths Lover

Ah ok, but I've heared that munkres treatment of algebraic topology is not that good. it's better to find another text (anyway, I don't intend to study algebraic topology currently)

@JasperLoy

@robjohn, If you're familiar woth munkres topology, Does the third chapter on connectedness and compactness require solid knowledge of metric spaces( two sections on chapter 2) ?

with*

robjohn

@MathsLover I've never read it

Maths Lover

12:05

Ah ok thanx :))

Jasper Loy

@MathsLover I recommend you Bredon's Topology and Geometry. It has good treatments of point set and algebraic topology, and treats differential topology too.

Maths Lover

Do you know anyone in the site who know the book so I can ask him\her ?@robjohn

Jasper Loy

@MathsLover Even if he reads the whole book, he cannot answer your question. It is kind of a non-question.

Maths Lover

@JasperLoy, I will give it a look(maybe after studying general topology from munkres?! )

Jasper Loy

@MathsLover Bredon is the only book that treats point set, differential and algebraic topology in some detail.

Maths Lover

12:09

Ah cool, @JasperLoy , I will look for it soon :)

Ok @jasper , I will move on to the next chapter. If I found I needs more on metric spaces I can go back to it and then go forward again , isn't a good idea ?

robjohn

12:32

@MathsLover I imagine there are quite a few, but I don't know of anyone in particular.

Balarka Sen

12:42

@blue of course it was I.

nullgeppetto

Hi people! Is it reasonable (or just not-crazy) to define a symmetric positive definite kernel on $\mathcal{X}\times\mathbb{S}_{++}^n times \mathcal{X}\times\mathbb{S}_{++}^n$, where $\mathcal{X}$ denotes a non-empty set, and $\mathbb{S}_{++}^n$ the space of symmetric positive definite $n\times n$ matrices? Thanks a lot! cc @robjohn, @DanielFischer ( P.s.: Is it ok to mention you, or it's just as annoying it seems to be? )

Jasper Loy

@MathsLover Yes, that is a good idea.

salbeira

To show that there is no local max or min in a multi-dimensional function at any point, I need to show that the hesse matrix at that point is indefinite, right?

Daniel Fischer

13:03

@nullgeppetto It is not-crazy. Whether it is reasonable (in the sense that it helps you come nearer to your goal) depends.

nullgeppetto

Thanks @DanielFischer! First it's good that it may make sense and it's not crazy..

Jasper Loy

@DanielFischer Have you thought of showing another picture, other than the one you are coughing?

nullgeppetto

@DanielFischer, actually I have just asked a question about, but I'm afraid that posting it here may be considered as spamming.. Should I post it?

Daniel Fischer

@JasperLoy I have thought about that, but haven't searched through everything to see whether I have one where I don't look too stupid.

Jasper Loy

@DanielFischer I would like to ask you if you have learnt about Banach manifolds or integrals of Banach-valued functions. They don't seem mainstream stuff.

Daniel Fischer

13:09

@nullgeppetto I have looked at it. I don't think I can answer it :(

nullgeppetto

Thanks @DanielFischer! No problem at all, thanks for taking a look!

Daniel Fischer

@JasperLoy A bit here and there.

Jasper Loy

@DanielFischer I hesitate to read Lang's Fundamentals of Differential Geometry and his Real and Functional Analysis because he uses them in those books.

Daniel Fischer

@JasperLoy What about just dipping in your toe and seeing whether it is to your liking?

Jasper Loy

@DanielFischer OK, after I have read the more mainstream books.

Chris's sis

13:24

@robjohn I'll show you my proof later on, I didn't put it on paper yet (it won't take too long anyway).

(everything flows naturally)

Alyosha

In the definition of a normal extension, '$K$ is a splitting field for some family of polynomials in $F[X]$', why not just say `$K$ is a splitting field for at least one polynomial in $F[X]$', since a single polynomial still forms a family of polynomials (with {1} being the index set)?

Am I right in saying the two are equivalent?

This is a sanity check.

Daniel Fischer

@Alyosha If $K$ is the splitting field of an infinite family of polynomials, it need not be the splitting field of a single polynomial.

If $K$ is the splitting field of $p$, then $[K:F] < \infty$.

Alyosha

@DanielFischer Thank you.

Jasper Loy

You are welcome.

barznjy

13:40

Hi,
I have the following integral;

$$y=\int_{0}^{\infty}\frac{e^{x f}}{m+x}\gamma\left(a,h x\right) dx$$
where $f,m, and h \in reals$
f,m, and h are $>$ 0

$a \in integers$, a=1,2,3,...
$\gamma\left(a,h x\right)$ is the lower incomplete gamma function

Can anyone help me how to solve it?
Thank you very much

Jasper Loy

13:52

Currently there are 12 accounts on math that say "please delete me" Can a mod delete them, or do the users need to have emailed the SE staff? @robjohn

Daniel Fischer

@JasperLoy One needs to email the staff.

(And how did you count them?)

Jasper Loy

@DanielFischer I just searched the usernames for "please delete me" Sorry, I was referring to the usernames, not the profiles.

2 of the 12 are mine. 1 I requested the SE staff to delete, and the other I did not.

Daniel Fischer

I see.

Jasper Loy

Anyway, no harm just leaving them there.

Balarka Sen

Hmm, weird.

whenever i press the downarrow key of the laptop keyboard, the slider descents all at once to the bottom of the page.

any idea why this is happening? just a key problem or is there something to it?

oh hello @TedShifrin

Jasper Loy

14:04

@BalarkaSen Did you press another button such as "fn" on a laptop?

Balarka Sen

@JasperLoy um, i am not sure. it ain't my machine.

Jasper Loy

@BalarkaSen Ah, then don't bother, lol.

Balarka Sen

it's just annoying.

Jasper Loy

I always thought that only girls liked to use the word annoying.

Balarka Sen

you were wrong.

ah, it's definitely a key problem. the num lock arrow keys are alright.

robjohn

14:11

@Chris'ssis It usually does when you know the idea to exploit.

Alyosha

@BalarkaSen Are all the roots of $\text{min}(\alpha, k)$ in $k(\alpha)$?

Jasper Loy

@Alyosha I believe you can just use \min.

Alyosha

@JasperLoy OK, thanks.

robjohn

14:30

@JasperLoy they need to contact the SE staff. The mods cannot do anything about that.

Sawarnik

@JasperLoy Orly?

robjohn

AFK for a bit... BBL

Balarka Sen

@Alyosha what is $\alpha$? what is $k$? what is $k(\alpha)$?

Looks like you you are talking of fields, but then the min doesn't make sense.

@Sawarnik Sherlock is awesome.

salbeira

How can I show that some set is a manifold?

Sawarnik

@BalarkaSen Oh awesome, you read the novels? Or BBC?

Balarka Sen

14:54

@Sawarnik Read a view short stories, but novels mostly.

Khallil

@Sawarnik In my opinion, the BBC is overrated. Match of the Day is the only thing worth watching.

Balarka Sen

The Hound of Baskerville was cool, but somewhat slow. A Study In Scarlet was excellent and now I am reading The Sign of The Four, which seems interesting also.

Sawarnik

@BalarkaSen Nice, but

the stories are much more interesting.

Do you have a book?

Balarka Sen

Yes. A collection.

Sawarnik

In 2 volumes right?

Balarka Sen

14:57

No, it's a better edition with all at once. Novels, short stories, returning of Holmes and some other stuffs I haven't peeked in.

I have read the final story though. It was cool.

Sawarnik

It must be a big book :|

Which story?

Balarka Sen

@Sawarnik Very big.

Sawarnik

@BalarkaSen I think Baskerville was the best of the novels. Scarlet was his first novel so its a bit different, you get the facts all at the end.

Balarka Sen

What d'you think of The Sign of Four?

Sawarnik

Average.

Which story?

Balarka Sen

15:00

What story?

Sawarnik

I think the last few ones are poor, but otherwise they are awesome.

@BalarkaSen The one you read.

1 min...

Balarka Sen

Oh, you mean the Final Problem?

I have read both A Scandal in Bohemia and A Case Of Identity as well.

Sawarnik

Back.

@BalarkaSen Nice.

They are good as well, Bohemia is the only one in which he is totally decieved :D

Balarka Sen

Yeah. The only one in which he failed.

Sawarnik

Read the Charles Augustus Something...that is something different as well :D

Balarka Sen

15:07

Will do.

I especially liked the analysis about the watch of Watson in the Sign of Four.

Sawarnik

@Khallil True. Season 3, especially episode 3 was very poor.

@BalarkaSen Don't remember .... :|

Balarka Sen

what the heck do you guys mean by BBC?

Sawarnik

@BalarkaSen Google :p

Balarka Sen

not interested.

Sawarnik

:/ Alright. Then why do you ask.

Balarka Sen

15:09

oh, by the way, @Sawarnik, Edgar Allan Poe is mentioned at Sherlock Holmes.

Sawarnik

@BalarkaSen I know :P

That why I ve 'heard' his name.

@BalarkaSen Speckled Band is a chilling story, the writer's favorite. Well are interesting!

@Khallil You ere?

Khallil

I'm here, @Sawarnik.

We're referring to the British Broadcasting Corporation, @BalarkaSen. (bbc.co.uk) (en.wikipedia.org/wiki/BBC)

Sawarnik

lol ... that was not what he was asking ... it was the connection between Sherlock and BBC that puzzled him

@Khallil Don't you think S2E3 was awesome?

Khallil

Oh! Silly me, I wasn't keeping up with conversation. I haven't seen any of Sherlock.

I'm saving the show so that I can 'marathon it', so to speak.

At the moment, I'm watching Yu Yu Hakusho, which has become my joint favourite show (along with ZOIDS)!

Sawarnik

@Khallil $$\huge{\text{-_-}}$$

Khallil

15:25

@Sawarnik This post, was brought to you by trial and error! ^_^

Sawarnik

@Khallil Job done, though.

@Khallil You have a lot of Indian friends on FB.

Khallil

I do?

Sawarnik

Yes.

Khallil

I guess the area I live in is predominantly Asian.

Not that I'm complaining. ;-)

Have you done any set theory, @Sawarnik?

Sawarnik

@Khallil Why the ;-)?

@Khallil Very basic ones.

Khallil

15:30

Me too. I've just started and it seems pretty cool. I still can't draw a nice diagram for the Cartesian product of three sets though ...

Sawarnik

@Khallil Oh damn, that would be in 3 dimensions :|

Khallil

No reason in particular. I like that winking face! ;-)

Exactly, the 2 dimensional one is easy, but doing it in three dimensions is a bit complicated.

It's definitely possible, but tedious and cumbersome all the same.

Sawarnik

@Khallil And how would you draw it on paper :| :|

Khallil

Not a clue. I'm thinking it'd look like sort of cube. Why the :| face?

Sawarnik

Because you say as if you have to draw it and you are not great at it ... :|

Khallil

15:33

I don't follow. Is it not possible to draw it on paper?

Sawarnik

I don't follow.

Is it easy to draw any kind of cuboid on a 2D paper?

Khallil

Huy

@DanielFischer: Do you have a moment? It's a simple question about $L^p$ spaces and Riesz' representation theorem.

Sawarnik

@Khallil Draw $\mathbb{A}\times\mathbb{Q}$ where $A$ is the set of irrationals.

:P

Khallil

That seems really hard.

Sawarnik

15:35

@Khallil With the coordinates and lengths and not cubes ... I can't do it. Do it if you want.

Martin Sleziak

@DanielFischer Dobrý deň! Do you speak Slovak or did you simply find translation somewhere. (BTW for some reason I thought you were from Austria; maybe because of Arthur Fischer. Judging by the discussion with sp about World Cup, it seems that you're from Germany.|

Sawarnik

@Khallil :P

You told 2D was easy. You can just fill up the whole paper :P

Khallil

Hahahahahaha!

Sawarnik

@Khallil Functions are interesting, yup.

Khallil

Have you seen the rectangular array for finding ordered pairs in a Cartesian product of two sets?

Sawarnik

15:37

Rect. array?

Khallil

Sawarnik

Yes so?

Khallil

I was thinking of doing that for three sets.

Sawarnik

@Khallil -_- Alright, can a function have an empty domain or empty codomain?

Khallil

Yea.

Actually, no.

Sawarnik

15:39

Like?

See the definition carefully and tell :P

Khallil

Each input has to relate to exactly one output value.

Sawarnik

But has there to be one?

It says about the existence of $b$ but not $a$?

Khallil

So does $a$ have to exist?

Sawarnik

@Khallil Enjoy this, chat.stackexchange.com/transcript/message/15460171#15460171

Khallil

You didn't phrase your question correctly!

May 11 at 11:32, by Sawarnik

Can a function have an nonempty domain or nonempty codomain?

You meant empty!

Sawarnik

15:45

@Khallil Oops.

Sorry.

@Khallil I correct it after 2 messages.

Khallil

No? =S

Sawarnik

Ok :|

Forget it .. :|

..bye

Khallil

Oh, saw it!

You're going?

Talk to you later. ^_^

Daniel Fischer

@MartinSleziak I was in Prague for a week in 1985 (Czechoslovakia just had won the Ice Hockey world championships, so everybody was very happy), and I remember a very little of Czech from then.

@Huy I think I now have a moment.

Martin Sleziak

@DanielFischer Czech and Slovak language are indeed very similar.

Huy

15:51

@DanielFischer: I'm looking at a proof for the following:
Let $1 < p \leq \infty$ and $u \in L^p(I)$ for an open interval $I$. Then, $u \in W^{1,p}(I)$ if and only if there is a $C \in \mathbb{R}$ such that $$| \int_I u \varphi' dx | \leq C \| \varphi \|_{L^q}$$ for all $\varphi \in C_c^\infty$ and $p^{-1} + q^{-1} = 1$.

Daniel Fischer

@MartinSleziak Indeed. And for a speaker of a non-slavic language, they seem even more similar than to Slavs, I'd think.

Huy

For the reverse direction, the proof begins as follows: If the inequality holds, we can continuously extend $$C_c^\infty(I) \ni \varphi \mapsto \int_I u \varphi' dx $$ to $l \in (L^q(I))^*$.

Daniel Fischer

@Huy Yup, Hahn-Banach.

Huy

Then, since $q < \infty$, there exists $g \in L^p$ such that $$l(\varphi) = \int_I g \varphi dx$$ for all $\varphi \in L^q$.

Daniel Fischer

Riesz, indeed.

Huy

15:54

I think this is Riesz', but I am a bit confused because is the scalar product of $L^2$ also the scalar product on $L^q$ for finite $q$?

Alyosha

@BalarkaSen $\alpha \in E, k \subseteq E$, then $\min(\alpha, k)$ is the minimal polynomial in $k[X]$ that has $\alpha$ as a root.

John Doe

Hi all

Daniel Fischer

@Huy There are a lot of (similar/related) theorems that go under the name of "Riesz representation theorem". Here we're dealing with the one describing $(L^p)^\ast$.

John Doe

Do you speak any Dutch? @DanielFischer

Daniel Fischer

@JohnDoe Hardly any. And I have a terrible accent. I can understand more, however.

Huy

15:56

@DanielFischer: I thought Riesz only holds for $L^2$ and its canonical inner product?

How exactly does it work for spaces which are non-Hilbert?

Daniel Fischer

@Huy That's one of the representation theorems, the one describing the dual space of a Hilbert space. The one describing $(L^q)^\ast$ for $1\leqslant q < \infty$ is another one (with nonempty intersection).

John Doe

@DanielFischer Oh okay, in South Africa we speak an offshoot of Dutch called Afrikaans...but it's quite similar.

Daniel Fischer

@JohnDoe Isn't English also an official language? And nowadays probably some Zulu or Xhosa dialects too?

Chris's sis

@robjohn

John Doe

@DanielFischer There are 11 official languages...Zulu and Xhosa are the mother tongue of the majority of population followed by Afrikaans and then English. But I guess English is mostly widely spoken across the board here as a second language.

Daniel Fischer

16:04

@JohnDoe It's the one that is most useful for international communication, so that's a strong incentive to learn it.

John Doe

@DanielFischer Yes for sure. Funny thing, our family spent a year in America when I was 9. So when we were irritated by the Americans we would swear at them in Afrikaans. But the thing that makes Afrikaans different from Dutch is that it has a mixture of many other languages as well, one of which is English.

So my brother tried to swear at some guys in Afrikaans but it sounded so much like the English version that they picked up on what he was saying. They didn't take it well but he convinced them that it meant something different in Afrikaans.

Daniel Fischer

I guess it would have been harder to wiggle out of that if one of the guys had been dutch ;)

John Doe

Yeah :)

Khallil

16:44

Hey, @Alizter.

Alizter

@Khallil hai

Khallil

How're you doing?

@Alizter

Alizter

@Khallil Good.

Fernando Martin

Hey guys, I have to prove that if $f\in L^1_{\text{loc}}$ and $\int fg dx=0$ for all $g\in C_0^\infty$, then $f=0$ ae

Khallil

That's good to hear! I'm really enjoying Hammond's Book of Proof. ^_^

Fernando Martin

16:57

I have a faulty argument involving approximations to the identity

anyone up to try to fix it? :)

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$Mathematics$ Mathematics